Find the Derivative - d/dx natural log of x+ square root of x^2+1

Problem

d(ln(x+√(,x2+1)))/d(x)

Solution

1. Identify the outer function as the natural logarithm and the inner function as u=x+√(,x2+1)

2. Apply the chain rule for the natural logarithm, which states d(ln(u))/d(x)=1/u⋅d(u)/d(x)

d(ln(x+√(,x2+1)))/d(x)=1/(x+√(,x2+1))⋅d(x+√(,x2+1))/d(x)

3. Differentiate the inner expression term by term.

d(x+√(,x2+1))/d(x)=1+d(x2+1)/d(x)

4. Apply the chain rule to the square root term.

d(x2+1)/d(x)=1/2*(x2+1)(−1/2)⋅2*x

5. Simplify the derivative of the inner expression.

1+x/√(,x2+1)=(√(,x2+1)+x)/√(,x2+1)

6. Substitute this back into the chain rule expression.

1/(x+√(,x2+1))⋅(x+√(,x2+1))/√(,x2+1)

7. Cancel the common factor x+√(,x2+1) from the numerator and denominator.

1/√(,x2+1)

Final Answer

d(ln(x+√(,x2+1)))/d(x)=1/√(,x2+1)

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